cp's OEIS Frontend

A165735 Surviving integers under the double-count Josephus problem (see A054995), modulo 3.

%I A165735 #28 Feb 16 2025 08:33:11
%S A165735 1,2,2,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,1,1,1,1,
%T A165735 1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,
%U A165735 2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
%N A165735 Surviving integers under the double-count Josephus problem (see A054995), modulo 3.
%C A165735 Old name was: The pattern is obvious. The sequence can be divided into subsequences of {1,1,1,...} and {2,2,2,...}.
%C A165735 Let n be a natural number. We put n numbers in a circle, and we are going to remove every third number. Let J3(n) be the last number that remains. This is the traditional Josephus Problem. Let J3 (mod 3) be the residue of the sequence J3(n) under mod 3. J3 (mod 3) produces the sequence {1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2,...}.
%H A165735 Hiroshi Matsui, Masakazu Naito and Naoyuki Totani, <a href="https://scholar.rose-hulman.edu/rhumj/vol10/iss1/9">The Period and the Distribution of the Fibonacci-like Sequence Under Various Moduli</a>, Undergraduate Math Journal, Rose-Hulman Institute of Technology, Vol. 10, Issue 1, 2009.
%H A165735 Masakazu Naito and Ryohei Miyadera, <a href="http://www.mi.sanu.ac.rs/vismath/miyadera2009April/JoseVis.html"> The Self-Similarity of the Josephus Problem and its Variants</a>, Visual Mathematics, Volume 11, No.2, 2009.
%H A165735 Wolfram MathWorld, <a href="https://mathworld.wolfram.com/JosephusProblem.html">Josephus Problem</a>
%H A165735 <a href="/index/J#Josephus">Index entries for sequences related to the Josephus Problem</a>
%F A165735 (1) J3(1) = 1 and J3(2) = 2.
%F A165735 (2) J3(3m) = J3(2m) + [(J3(2m)-1)/2].
%F A165735 (3a) J3(3m+1) = 3m + 1 (if J3(2m + 1) = 1).
%F A165735 (3b) J3(3m+1) = J3(2m+1) + [J3(2m+1)/2] - 2 (if J3(2m + 1) > 1).
%F A165735 (4) J3(3m+2) = J3(2m+1) + [J3(2m+1)/2] + 1
%F A165735 a(n) = A010872(A054995(n)). - _Gordon Atkinson_, Aug 21 2019
%e A165735 If we use n = 10, then we put numbers 1,2,3,4,5,6,7,8,9,10 in a circle. We eliminate 3,6,9,2,7,1,8,5,10, and the last number that remains is 4. Therefore J3(10) = 4 and J3(10) = 1 mod 3.
%t A165735 J3[1] = 1; J3[2] = 2; J3[n_] := J3[n] = Block[{m, t}, t = Mod[n, 3]; m = (n - t)/3; Which[t == 0, J3[2 m] + Floor[(J3[2 m] - 1)/2], t == 1, If[J3[2 m + 1] == 1, 3 m + 1, J3[2 m + 1] + Floor[J3[2 m + 1]/2] - 2], t == 2, J3[2 m + 1] + Floor[J3[2 m + 1]/2] + 1]]; Table[Mod[J3[n], 3], {n, 1, 200}]
%Y A165735 Cf. A010872, A054995, A114144, A113648, A165556.
%K A165735 nonn
%O A165735 1,2
%A A165735 _Ryohei Miyadera_ and Masakazu Naito, Sep 25 2009
%E A165735 New name from _Gordon Atkinson_, Aug 21 2019